Simplify the following expression and state the condition under which the simplification is valid. You can assume that $x \neq 0$. $k = \dfrac{27x - 63}{-6} \times \dfrac{5x}{18x^2 - 42x} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $k = \dfrac{ (27x - 63) \times 5x } { -6 \times (18x^2 - 42x) } $ $ k = \dfrac {5x \times 9(3x - 7)} {-6 \times 6x(3x - 7)} $ $ k = \dfrac{45x(3x - 7)}{-36x(3x - 7)} $ We can cancel the $3x - 7$ so long as $3x - 7 \neq 0$ Therefore $x \neq \dfrac{7}{3}$ $k = \dfrac{45x \cancel{(3x - 7})}{-36x \cancel{(3x - 7)}} = -\dfrac{45x}{36x} = -\dfrac{5}{4} $